A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

No. of defective tubes |
Less than 5 | 5 – 9 | 10 – 14 | 15 – 9 | 20 – 24 | 25 – 29 | 30 and above |

No. of tubes |
45 | 51 | 84 | 39 | 20 | 8 | 4 |

Estimate the number of defective tubes in the central batch.

#### Solution

To find the number of defective tubes in the central batch, we have to find Q_{2}.

Since, the given data is not continuous, we have to convert it in the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.

∴ the class intervals will be

Less than 4.5, 4.5 – 9.5, etc.

We construct the less than cumulative frequency table as given below:

No. of defective tubes |
No. of tubes (f) |
Less than cumulative frequency (c.f.) |

Less than 4.5 | 45 | 45 |

4.5 – 9.5 | 51 | 96 |

9.5 – 14.5 | 84 | 180 ← Q_{2} |

14.5 – 19.5 | 39 | 219 |

19.5 – 24.5 | 20 | 239 |

24.5 – 29.5 | 8 | 247 |

29.5 and above | 4 | 251 |

Total |
251 |

Here, N = 251

Q_{2} class = class containing `((2"N")/4)^"th"` observation

∴ `(2"N")/4=(2xx251)/4` = 125.5

Cumulative frequency which is just greater than (or equal to) 125.5 is 180.

∴ Q_{2} lies in the class 9.5 – 14.5

∴ L = 9.5, h = 5, f = 84, c.f. = 96

∴ Q_{2} = `"L" "h"/"f"((2"N")/4-"c.f.")`

= `9.5 + 5/84 (125.5 - 96)`

= `9.5 + 5/84 xx 29.5`

= `9.5 + (147.5)/84`

= 9.5 + 1.76

= 11.26